If longest wavelength for Paschen series in H-atom is α/7R. Find out α

Answer :

Correct answer is : 144

$\frac{1}{\mathrm{\lambda \; <!--\; \lambda \; -->}}=R(1{)}^2(\frac{1}{(3{)}^2}-\; <!--\; -\; -->\frac{1}{(4{)}^2})$

$\frac{1}{\mathrm{\lambda \; <!--\; \lambda \; -->}}=R(\frac{1}{9}-\; <!--\; -\; -->\frac{1}{16})$

$\frac{1}{\mathrm{\lambda \; <!--\; \lambda \; -->}}=R(\frac{7}{9\times \; <!--\; \times \; -->16})$

$\mathrm{\lambda \; <!--\; \lambda \; -->}=\frac{9\times \; <!--\; \times \; -->16}{7R}$ = $\frac{144}{7R}$